Advertisements
Advertisements
Question
The sum of three numbers in A.P. is 3 and their product is – 35. Find the numbers.
Solution
Sum of three numbers which are in A.P. = 3
Their product = -35
Let three numbers which are in A.P.
a – d, a, a + d
a – d + a + a + d = 3
⇒ 3a = 3,
⇒ a = `(3)/(3)` = 1
and
(a – d) x a (a + d) = – 35
(1 – d) x 1 x (1 + d) = – 35
12 – d2 = – 35
1 – d2 = – 35
⇒ d2 = 35 + 1 = 36
∴ d = ±6
If d = 6
∴ Number are 1 – 6, 1, 1 + 6
= – 5, 1, 7
If d = – 6
1 + 6, 1, 1 – 6
⇒ 7, 1, – 5
Hence numbers on A.P. are
– 5, 1, 7 or 7, 1, – 5.
APPEARS IN
RELATED QUESTIONS
The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.
Q.1
Find the general term (nth term) and 23rd term of the sequence 3, 1, –1, –3, ........... .
If the nth term of the A.P. 58, 60, 62, .... is equal to the nth term of the A.P. –2, 5, 12, …., find the value of n.
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31st term of this A.P.
Find the A.P. whose nth term is 7 – 3K. Also find the 20th term.
Find the indicated terms in each of following A.P.s: 1, 6, 11, 16, …; a20
If the common difference of an A.P. is – 3 and the 18th term is – 5, then find its first term.
Determine the A.P. whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Find the 31st term of an A.P. whose 11th term is 38 and 6th term is 73.